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Proof step 8 from proof oc-filt (previous step)

a ∈ y, ∀ a, y. a ∈ y → #oc(a, filter(x0, y)) = #oc(a, x0) ⊦ #oc(a, filter(a0 ' + x0, y)) = #oc(a, a0 ') + #oc(a, x0), a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ y, ∀ a, y. a ∈ y → #oc(a, filter(x0, y)) = #oc(a, x0) ⊦ #oc(a, filter(a0 ' + x0, y)) = #oc(a, x0), a0 = a

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)